--- comments: true --- # 15.2 Fractional knapsack problem !!! question Given $n$ items, the weight of the $i$-th item is $wgt[i-1]$ and its value is $val[i-1]$, and a knapsack with a capacity of $cap$. Each item can be chosen only once, **but a part of the item can be selected, with its value calculated based on the proportion of the weight chosen**, what is the maximum value of the items in the knapsack under the limited capacity? An example is shown in Figure 15-3. ![Example data of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_example.png){ class="animation-figure" }
Figure 15-3 Example data of the fractional knapsack problem
The fractional knapsack problem is very similar overall to the 0-1 knapsack problem, involving the current item $i$ and capacity $c$, aiming to maximize the value within the limited capacity of the knapsack. The difference is that, in this problem, only a part of an item can be chosen. As shown in Figure 15-4, **we can arbitrarily split the items and calculate the corresponding value based on the weight proportion**. 1. For item $i$, its value per unit weight is $val[i-1] / wgt[i-1]$, referred to as the unit value. 2. Suppose we put a part of item $i$ with weight $w$ into the knapsack, then the value added to the knapsack is $w \times val[i-1] / wgt[i-1]$. ![Value per unit weight of the item](fractional_knapsack_problem.assets/fractional_knapsack_unit_value.png){ class="animation-figure" }Figure 15-4 Value per unit weight of the item
### 1. Greedy strategy determination Maximizing the total value of the items in the knapsack **essentially means maximizing the value per unit weight**. From this, the greedy strategy shown in Figure 15-5 can be deduced. 1. Sort the items by their unit value from high to low. 2. Iterate over all items, **greedily choosing the item with the highest unit value in each round**. 3. If the remaining capacity of the knapsack is insufficient, use part of the current item to fill the knapsack. ![Greedy strategy of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_greedy_strategy.png){ class="animation-figure" }Figure 15-5 Greedy strategy of the fractional knapsack problem
### 2. Code implementation We have created an `Item` class in order to sort the items by their unit value. We loop and make greedy choices until the knapsack is full, then exit and return the solution: === "Python" ```python title="fractional_knapsack.py" class Item: """Item""" def __init__(self, w: int, v: int): self.w = w # Item weight self.v = v # Item value def fractional_knapsack(wgt: list[int], val: list[int], cap: int) -> int: """Fractional knapsack: Greedy""" # Create an item list, containing two properties: weight, value items = [Item(w, v) for w, v in zip(wgt, val)] # Sort by unit value item.v / item.w from high to low items.sort(key=lambda item: item.v / item.w, reverse=True) # Loop for greedy selection res = 0 for item in items: if item.w <= cap: # If the remaining capacity is sufficient, put the entire item into the knapsack res += item.v cap -= item.w else: # If the remaining capacity is insufficient, put part of the item into the knapsack res += (item.v / item.w) * cap # No remaining capacity left, thus break the loop break return res ``` === "C++" ```cpp title="fractional_knapsack.cpp" [class]{Item}-[func]{} [class]{}-[func]{fractionalKnapsack} ``` === "Java" ```java title="fractional_knapsack.java" /* Item */ class Item { int w; // Item weight int v; // Item value public Item(int w, int v) { this.w = w; this.v = v; } } /* Fractional knapsack: Greedy */ double fractionalKnapsack(int[] wgt, int[] val, int cap) { // Create an item list, containing two properties: weight, value Item[] items = new Item[wgt.length]; for (int i = 0; i < wgt.length; i++) { items[i] = new Item(wgt[i], val[i]); } // Sort by unit value item.v / item.w from high to low Arrays.sort(items, Comparator.comparingDouble(item -> -((double) item.v / item.w))); // Loop for greedy selection double res = 0; for (Item item : items) { if (item.w <= cap) { // If the remaining capacity is sufficient, put the entire item into the knapsack res += item.v; cap -= item.w; } else { // If the remaining capacity is insufficient, put part of the item into the knapsack res += (double) item.v / item.w * cap; // No remaining capacity left, thus break the loop break; } } return res; } ``` === "C#" ```csharp title="fractional_knapsack.cs" [class]{Item}-[func]{} [class]{fractional_knapsack}-[func]{FractionalKnapsack} ``` === "Go" ```go title="fractional_knapsack.go" [class]{Item}-[func]{} [class]{}-[func]{fractionalKnapsack} ``` === "Swift" ```swift title="fractional_knapsack.swift" [class]{Item}-[func]{} [class]{}-[func]{fractionalKnapsack} ``` === "JS" ```javascript title="fractional_knapsack.js" [class]{Item}-[func]{} [class]{}-[func]{fractionalKnapsack} ``` === "TS" ```typescript title="fractional_knapsack.ts" [class]{Item}-[func]{} [class]{}-[func]{fractionalKnapsack} ``` === "Dart" ```dart title="fractional_knapsack.dart" [class]{Item}-[func]{} [class]{}-[func]{fractionalKnapsack} ``` === "Rust" ```rust title="fractional_knapsack.rs" [class]{Item}-[func]{} [class]{}-[func]{fractional_knapsack} ``` === "C" ```c title="fractional_knapsack.c" [class]{Item}-[func]{} [class]{}-[func]{fractionalKnapsack} ``` === "Kotlin" ```kotlin title="fractional_knapsack.kt" [class]{Item}-[func]{} [class]{}-[func]{fractionalKnapsack} ``` === "Ruby" ```ruby title="fractional_knapsack.rb" [class]{Item}-[func]{} [class]{}-[func]{fractional_knapsack} ``` === "Zig" ```zig title="fractional_knapsack.zig" [class]{Item}-[func]{} [class]{}-[func]{fractionalKnapsack} ``` Apart from sorting, in the worst case, the entire list of items needs to be traversed, **hence the time complexity is $O(n)$**, where $n$ is the number of items. Since an `Item` object list is initialized, **the space complexity is $O(n)$**. ### 3. Correctness proof Using proof by contradiction. Suppose item $x$ has the highest unit value, and some algorithm yields a maximum value `res`, but the solution does not include item $x$. Now remove a unit weight of any item from the knapsack and replace it with a unit weight of item $x$. Since the unit value of item $x$ is the highest, the total value after replacement will definitely be greater than `res`. **This contradicts the assumption that `res` is the optimal solution, proving that the optimal solution must include item $x$**. For other items in this solution, we can also construct the above contradiction. Overall, **items with greater unit value are always better choices**, proving that the greedy strategy is effective. As shown in Figure 15-6, if the item weight and unit value are viewed as the horizontal and vertical axes of a two-dimensional chart respectively, the fractional knapsack problem can be transformed into "seeking the largest area enclosed within a limited horizontal axis range". This analogy can help us understand the effectiveness of the greedy strategy from a geometric perspective. ![Geometric representation of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_area_chart.png){ class="animation-figure" }Figure 15-6 Geometric representation of the fractional knapsack problem